
% \section*{Appendix}
\appendix

\noindent\underline{\em Proof of Theorem~\ref{thm:palpha}.} 
For each of $\gamma_{i},$ $\gamma_{ij},$ and $\gamma_{jk},$ we find that
$\tilde{U}_{i}(\bm{x};\gamma_{i})=0,$
$\tilde{U}_{i}(\bm{x};\gamma_{ij})=\tilde{U}_{j}(\bm{x};\gamma_{ij}),$ and
$\tilde{U}_{j}(\bm{x};\gamma_{jk})=\tilde{U}_{k}(\bm{x};\gamma_{jk}).$
Thus, at ${\bm p'}$, the equilibrium exists with $\bm{\alpha}$ since $T_{i}\le T_{j}\le T_{k}.$
For the given $\bm \alpha$, $p_l$, where $l \in \{ m,o,c\},$ can take other values only when $\alpha_l=0.$ 
\QED

\vspace{1cm}
\smallskip
\noindent\underline{\em Proof of Theorem~\ref{thm:pvolume}.}
% This proof is for the case $x^{O} \le x^{C}$. In the other case, the
% same proof is valid under the interchange of $o$ and $c$. 
 At the $p_i$ and $p_j$, $\tilde{U_{i}}({\bm
  x};\gamma_{ij})=\tilde{U_{j}}({\bm x};\gamma_{ij})$.  Moreover, the relationship $\tilde{U_{i}}({\bm
  x};\gamma)>\tilde{U_{j}}({\bm x};\gamma)$ holds when $\gamma <
\gamma_{ij},$ and the relationship $\tilde{U_{i}}({\bm
  x};\gamma)<\tilde{U_{j}}({\bm x};\gamma)$ holds when $\gamma >
\gamma_{ij}.$ Thus, $\tilde{U_{i}}({\bm x};\gamma)$ and
$\tilde{U_{j}}({\bm x};\gamma)$ intersect uniquely at $\gamma_{ij}$. 

The utility difference between service type $i$ and {\em mobile-only} service is given by: 
\begin{equation} 
 % \tilde{U_{o}}({\bm x};\gamma)-\tilde{U_{m}}({\bm
 %  x};\gamma)= 
 %  \pi_{o}^{O}\big(\gamma(x^{O})^{\theta}-\gamma(x^{M}(\gamma))^{\theta}+p_{v}^{M}x^{M}(\gamma)\big)-p_{o}.
\tilde{U_{i}}({\bm x};\gamma)-\tilde{U_{m}}({\bm
  x};\gamma)= 
  \gamma  \sum_{L \in  \{O,C \}} (x^L)^\theta \pi_i^L- (1-\pi_i^M)\big(\gamma(x^{M}(\gamma))^{\theta}-p_{v}^{M}x^{M}(\gamma)\big)-p_{i}.
\end{equation}
For $\gamma_b < \gamma_a$, we have:
\begin{eqnarray}
  \lefteqn{\tilde{U_{i}}({\bm x};\gamma_{a})-\tilde{U_{m}}({\bm
      x};\gamma_{a})+p_{i}}\\ \nonumber
  & = & \gamma_a  \sum_{L \in  \{O,C \}} (x^L)^\theta \pi_i^L-  (1-\pi_i^M)\big(\gamma_a
 (x^{M}(\gamma_a))^{\theta}-p_{v}^{M}x^{M}(\gamma_a)\big)\\ \nonumber
  & \ge & \gamma_b  \sum_{L \in  \{O,C \}} (x^L)^\theta \pi_i^L-   (1-\pi_i^M)\big(\gamma_b
 (x^{M}(\gamma_a))^{\theta}-p_{v}^{M}x^{M}(\gamma_a)\big)\label{eq:in}\\ \nonumber
  & \ge &
\gamma_b  \sum_{L \in  \{O,C \}} (x^L)^\theta \pi_i^L- (1-\pi_i^M)\big(\gamma_b  (x^{M}(\gamma_b))^{\theta}-p_{v}^{M}x^{M}(\gamma_b)\big)\label{eq:op}\\
  & = & \tilde{U_{o}}({\bm x};\gamma_{b})-\tilde{U_{m}}({\bm x};\gamma_{b})+p_{o},
\end{eqnarray}
where Eq. (\ref{eq:in}) is obtained from the condition $\min \{
x^{O},x^{C} \} \ge
x^{M}(\gamma_{a})$ and Eq. (\ref{eq:op}) is obtained from the fact that
$x^{M}(\gamma_{b})$ maximizes $\gamma_{b}x^{\theta}-p_{m}^{V}x.$ Thus,
$\tilde{U_{i}}({\bm x};\gamma)-\tilde{U_{m}}({\bm x};\gamma) $ is
increasing in $\gamma,$ and $\gamma_{mi}$ and $\gamma_{ij}$ are uniquely
determined with $\gamma_{ij}=1-\alpha_{j}$ and $\gamma_{mi}=\gamma_{ij}-\alpha_i.$

Assume that the capacity for macro BSs is not constrained.  
Remarking that:
\begin{equation}
\frac{dU_{l}({\bm
    x};\gamma)}{dx^{M}}=\pi_{l}^{M}\big(\theta\gamma(x^{M})^{\theta-1}-p_{v}^{M}\big), \qquad
l \in \{m,o,c \}
\end{equation}
the service rate of macro BSs for users with type $\gamma$ is
$x^{M}(\gamma)=(\frac{\theta\gamma}{p_{v}^{M}})^{\frac{1}{1-\theta}}.$
We also find that:
% \begin{eqnarray*}
%   \frac{dR}{dp_{v}^{M}} & = & N\theta^{\frac{1}{1-\theta}}(p_{v}^{M})^{\frac{1-2\theta}{1-\theta}}\Big\{\pi_{o}^{O}(\gamma_{mo})^{\frac{1}{1-\theta}}(1-\gamma_{mo})\\
%   &  & -\frac{\theta}{2-\theta}(\pi_{o}^{O}(\gamma_{mo})^{\frac{2-\theta}{1-\theta}}+1-\pi_{o}^{O})\Big\}.\end{eqnarray*}
\begin{equation}
  \frac{dR}{dp_{v}^{M}} =
  N\theta^{\frac{1}{1-\theta}}(p_{v}^{M})^{\frac{1-2\theta}{1-\theta}}\Big\{(1-\pi_{i}^{M})(\gamma_{mi})^{\frac{1}{1-\theta}}(1-\gamma_{mi})
  -\frac{\theta}{2-\theta}((1-\pi_{i}^{M})(\gamma_{mi})^{\frac{2-\theta}{1-\theta}}+\pi_{i}^{M})\Big\}.\end{equation}


Thus, when the condition (\ref{con:p}) is met, the revenue increases
with $p_{v}^{M}.$ Otherwise, owing to the capacity constraint, the revenue is maximized when $p_{v}^{M}$ satisfies following:
% \begin{eqnarray*}
%   C_{M} & = & \int\pi_{j(\gamma)}^{M}(\frac{\theta\gamma}{p_{v}^{M}})^{\frac{1}{1-\theta}}d\gamma\\
%   & = &
%   N\Big(\frac{1-\theta}{2-\theta}\Big)\Big(\pi_{o}^{O}(\gamma_{mo})^{\frac{2-\theta}{1-\theta}}+1-\pi_{o}^{O}\Big)\Big(\frac{\theta}{p_{v}^{M}}\Big)^{\frac{1}{1-\theta}}.
% \end{eqnarray*}
\begin{eqnarray*} 
  C_{M} &=&
  \int\pi_{l^*(\gamma)}^{M}(\frac{\theta\gamma}{p_{v}^{M}})^{\frac{1}{1-\theta}}d\gamma
  \\ &= &
  N\Big(\frac{1-\theta}{2-\theta}\Big)\Big((1-\pi_{i}^{M}) (\gamma_{mi})^{\frac{2-\theta}{1-\theta}}+\pi_{i}^{M}\Big)\Big(\frac{\theta}{p_{v}^{M}}\Big)^{\frac{1}{1-\theta}}.
\end{eqnarray*}
This concludes the proof. \QED
%\begin{lemma}
%Let $\alpha$ be the subscription ratio for the system. When there are only macro BSs, the operator get the maximum revenue when  $\alpha$ is $(1-\theta)/(2-\theta),$ in flat price.
%\label{lem:macro}
%\end{lemma}
%\begin{proof}
%Revenue $R$ can be represented as the function of $\alpha$:
%\[
%R=(1-\alpha)(\frac{C_{M}}{\alpha N})^{\theta}\alpha N.\]
%$R$ is differentiated as 
%\begin{eqnarray}
%\frac{dR}{d\alpha} & = & N(\frac{C_{M}}{N})^{\theta}\alpha^{-\theta}(1-\theta-(2-\theta)\alpha)\label{eq:first}\\
%\frac{d^{2}R}{d\alpha^{2}} & = & -N(\frac{C_{M}}{N})^{\theta}(1-\theta)(2+\theta(1-\alpha)\alpha^{-1}).\label{eq:second}\end{eqnarray}
%Since Equation~(\ref{eq:second}) is non-positive when $\alpha>0$ and $\theta \in [0,1],$ $R$ is a concave function of $\alpha.$ Thus, when Equation~(\ref{eq:first}) is zero, $\alpha=(1-\theta)/(2-\theta)$, $R$ is at the maximum.
%\end{proof}

\vspace{1cm}
 \medskip

\noindent\underline{\em Proof of Theorem~\ref{thm:femto}.}
We prove the theorem by finding the conditions on the femto costs for
{\em open-to-all} and {\em open-to-femto}, such that revenue when
there are only {\em mobile-only} users increases along with the change from a
mobile-only user to a femto user. 

Initially, let $\alpha$ be the subscription ratio for the system with only
{\em mobile-only} users. Subsequently, by jointly solving (\ref{eq:provider})
and (\ref{eq:user_flat}), $R$ in (\ref{eq:revenueflat}) is given by: 
$R=(1-\alpha)(C_{M}/(\alpha N))^{\theta}\alpha N,$
where we get:
 \begin{eqnarray}
   \frac{dR}{d\alpha} & = & N\Big(\frac{C_{M}}{N}\Big)^{\theta}\alpha^{-\theta}(1-\theta-(2-\theta)\alpha),\label{eq:first}\\
   \frac{d^{2}R}{d\alpha^{2}} & = & -N\Big(\frac{C_{M}}{N}\Big)^{\theta}(1-\theta)(2+\theta(1-\alpha)\alpha^{-1}).\label{eq:second}\end{eqnarray}
 Since Eq.~(\ref{eq:second}) is non-positive for $\alpha>0,\ \theta \in
 [0,1],$ $R$ is  concave in $\alpha.$ 
 Thus,  $R$ is maximized when 
 $\alpha^{*}:=\frac{1-\theta}{2-\theta}.$ Let $\gamma_{m}^{b}$ be
 $1-\alpha^{*}.$ Then, when no femto BS exists,
 according to Theorem~\ref{thm:palpha}, the maximum revenue is 
 $U_{m}({\bm x}^{(0)};\gamma_{m}^{b}) \alpha^{*}N$, where ${\bm x}^{(0)}$ is a traffic rate vector.

 Let us assume that, under the {\em open-to-all} policy, a user, whose
 type value is $\bar{\gamma}$, changes his or her
 service to \emph{open-femto}. Then, there are $\alpha^{*}N-1$
 \emph{mobile-only} users and an \emph{open-femto} user. From Theorem~\ref{thm:palpha}, the prices for \emph{mobile-only} and
 \emph{open-femto} are defined as
 \begin{eqnarray}
 p_{m}^{(1)} & := & U_{m}({\bm x}^{(1)};\gamma_{m}^{b}),\\
 p_{o}^{(1)} & := & U_{o}({\bm x}^{(1)};1)-U_{m}({\bm x}^{(1)};1)+U_{m}({\bm x}^{(1)};\gamma_{m}^{b}),
\end{eqnarray}
where  ${\bm x}^{(1)}$ is a traffic rate vector when there are
$\alpha^{*}N$ subscribers and they are \emph{mobile-only} users except
one subscribing to the \emph{open-femto} service.
 Thus, the revenue increment by introducing an \emph{open-femto} user is computed as
 % \begin{eqnarray*}
 % R^{*} & = & (U_{m}({\bm x}^{(1)};\gamma_{m}^{b})-U_{m}({\bm x}^{(0)};\gamma_{m}^{b}))\alpha^{*}N\\
 %  &  & +(U_{o}({\bm x}^{(1)};1)-U_{m}({\bm
 %  x}^{(1)};1))-c.\end{eqnarray*}
 \begin{equation}
 R^{*} = (U_{m}({\bm x}^{(1)};\gamma_{m}^{b})-U_{m}({\bm
   x}^{(0)};\gamma_{m}^{b}))\alpha^{*}N +(U_{o}({\bm x}^{(1)};1)-U_{m}({\bm x}^{(1)};1))-c.\end{equation}
 If $R^{*}$ is positive, the provider can gain more revenue with an open
 femto BS. Let the bound $\bar{c}_{O}$ denote the maximum femto
 cost that guarantees more profit with an open femto BS. Then the
 bound $\bar{c}_{O}$ is represented as
\begin{multline} 
\bar{c}_{O} = ((1-\alpha^{*})\alpha^{*}N\delta_{o}\beta+\delta_{i})(\frac{C_{O}}{1+\delta_{i}+
\delta_{o}\beta\alpha^{*}N})^{\theta}- \\ (1-\alpha^{*})\alpha^{*}N(\frac{C_{M}}{1+\alpha^{*}N})^{\theta}+((1-\alpha^{*})\alpha^{*}N(1-\delta_{o}\beta)-\delta_{i})(\frac{C_{M}}{1+\delta_{i}(\alpha^{*}N-1)+\delta_{o}(1-\beta)\alpha^{*}N})^{\theta}.
\end{multline}
Similarly, we can also compute the bound $\bar{c}_{C}$ which is the
maximum femto cost where a
provider gains revenue by introducing one {\em closed-femto} user.
\begin{equation}
\bar{c}_{C} = \delta_{i}C_{C}^{\theta}+((1-\alpha^{*})\alpha^{*}N-\delta_{i})(\frac{C_{M}}{1+\alpha^{*}N-\delta_{i}})^{\theta}-(1-\alpha^{*})\alpha^{*}N(\frac{C_{M}}{1+\alpha^{*}N})^{\theta}.
\end{equation}

Thus, under the {\em open-to-all} policy, $\bar{c}_{A}=\max
(\bar{c}_O,\bar{c}_C)$. Moreover, under the {\em open-to-femto} policy,
$\bar{c}_F=\bar{c}_C,$ 
because only one {\em open-femto} user has the same characteristic as
only one {\em closed-femto} user. 
Finally, we can conclude that $\bar{c}_A \ge \bar{c}_F.$ This concludes the proof. \QED



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